TSTP Solution File: SWV132+1 by SOS---2.0
View Problem
- Process Solution
%------------------------------------------------------------------------------
% File : SOS---2.0
% Problem : SWV132+1 : TPTP v8.1.0. Bugfixed v3.3.0.
% Transfm : none
% Format : tptp:raw
% Command : sos-script %s
% Computer : n027.cluster.edu
% Model : x86_64 x86_64
% CPU : Intel(R) Xeon(R) CPU E5-2620 v4 2.10GHz
% Memory : 8042.1875MB
% OS : Linux 3.10.0-693.el7.x86_64
% CPULimit : 300s
% WCLimit : 600s
% DateTime : Wed Jul 20 21:37:07 EDT 2022
% Result : Theorem 0.80s 1.04s
% Output : Refutation 0.80s
% Verified :
% SZS Type : -
% Comments :
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.04/0.13 % Problem : SWV132+1 : TPTP v8.1.0. Bugfixed v3.3.0.
% 0.04/0.14 % Command : sos-script %s
% 0.13/0.35 % Computer : n027.cluster.edu
% 0.13/0.35 % Model : x86_64 x86_64
% 0.13/0.35 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.13/0.35 % Memory : 8042.1875MB
% 0.13/0.35 % OS : Linux 3.10.0-693.el7.x86_64
% 0.13/0.36 % CPULimit : 300
% 0.13/0.36 % WCLimit : 600
% 0.13/0.36 % DateTime : Thu Jun 16 05:38:50 EDT 2022
% 0.13/0.36 % CPUTime :
% 0.80/1.04
% 0.80/1.04 -------- PROOF --------
% 0.80/1.04 % SZS status Unsatisfiable
% 0.80/1.04 % SZS output start Refutation
% 0.80/1.04 ----- Otter 3.2, August 2001 -----
% 0.80/1.04 The process was started by sandbox on n027.cluster.edu,
% 0.80/1.04 Thu Jun 16 05:38:50 2022
% 0.80/1.04 The command was "./sos". The process ID is 12162.
% 0.80/1.04
% 0.80/1.04 set(prolog_style_variables).
% 0.80/1.04 set(auto).
% 0.80/1.04 dependent: set(auto1).
% 0.80/1.04 dependent: set(process_input).
% 0.80/1.04 dependent: clear(print_kept).
% 0.80/1.04 dependent: clear(print_new_demod).
% 0.80/1.04 dependent: clear(print_back_demod).
% 0.80/1.04 dependent: clear(print_back_sub).
% 0.80/1.04 dependent: set(control_memory).
% 0.80/1.04 dependent: assign(max_mem, 12000).
% 0.80/1.04 dependent: assign(pick_given_ratio, 4).
% 0.80/1.04 dependent: assign(stats_level, 1).
% 0.80/1.04 dependent: assign(pick_semantic_ratio, 3).
% 0.80/1.04 dependent: assign(sos_limit, 5000).
% 0.80/1.04 dependent: assign(max_weight, 60).
% 0.80/1.04 clear(print_given).
% 0.80/1.04
% 0.80/1.04 formula_list(usable).
% 0.80/1.04
% 0.80/1.04 SCAN INPUT: prop=0, horn=0, equality=1, symmetry=0, max_lits=8.
% 0.80/1.04
% 0.80/1.04 This ia a non-Horn set with equality. The strategy will be
% 0.80/1.04 Knuth-Bendix, ordered hyper_res, ur_res, factoring, and
% 0.80/1.04 unit deletion, with positive clauses in sos and nonpositive
% 0.80/1.04 clauses in usable.
% 0.80/1.04
% 0.80/1.04 dependent: set(knuth_bendix).
% 0.80/1.04 dependent: set(para_from).
% 0.80/1.04 dependent: set(para_into).
% 0.80/1.04 dependent: clear(para_from_right).
% 0.80/1.04 dependent: clear(para_into_right).
% 0.80/1.04 dependent: set(para_from_vars).
% 0.80/1.04 dependent: set(eq_units_both_ways).
% 0.80/1.04 dependent: set(dynamic_demod_all).
% 0.80/1.04 dependent: set(dynamic_demod).
% 0.80/1.04 dependent: set(order_eq).
% 0.80/1.04 dependent: set(back_demod).
% 0.80/1.04 dependent: set(lrpo).
% 0.80/1.04 dependent: set(hyper_res).
% 0.80/1.04 dependent: set(unit_deletion).
% 0.80/1.04 dependent: set(factor).
% 0.80/1.04
% 0.80/1.04 ------------> process usable:
% 0.80/1.04
% 0.80/1.04 ------------> process sos:
% 0.80/1.04
% 0.80/1.04 -----> EMPTY CLAUSE at 0.65 sec ----> 913 [copy,912,unit_del,240,247] {-} $F.
% 0.80/1.04
% 0.80/1.04 Length of proof is 0. Level of proof is 0.
% 0.80/1.04
% 0.80/1.04 ---------------- PROOF ----------------
% 0.80/1.04 % SZS status Theorem
% 0.80/1.04 % SZS output start Refutation
% 0.80/1.04
% 0.80/1.04 240 [] {-} -leq(tptp_float_0_001,pv1341).
% 0.80/1.04 247 [] {-} -leq(s_best7,n3).
% 0.80/1.04 912 [] {-} leq(tptp_float_0_001,pv1341)|leq(s_best7,n3).
% 0.80/1.04 913 [copy,912,unit_del,240,247] {-} $F.
% 0.80/1.04
% 0.80/1.04 % SZS output end Refutation
% 0.80/1.04 ------------ end of proof -------------
% 0.80/1.04
% 0.80/1.04
% 0.80/1.04 Search stopped by max_proofs option.
% 0.80/1.04
% 0.80/1.04
% 0.80/1.04 Search stopped by max_proofs option.
% 0.80/1.04
% 0.80/1.04 ============ end of search ============
% 0.80/1.04
% 0.80/1.04 That finishes the proof of the theorem.
% 0.80/1.04
% 0.80/1.04 Process 12162 finished Thu Jun 16 05:38:51 2022
%------------------------------------------------------------------------------